Aboxer punches a sheet of paper in mid air, and brings it from rest up to a speed of 25 m/s in 0.05 s. if the mass of the paper is 0.005 kg, what force does the boxer exert on it

F = 2.5 N

Explanation:

Because the sheet of paper moves with a uniformly accelerated movement, we apply the following formulas:

vf= v₀+at Formula (1)

Where:  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

We apply Newton's second law:

F=m*a Formula (2)

F ; Force (N)

m: mass  (kg)

a: acceleration  (m/s² )

Problem development

We calculate the acceleration using the formula (1):

vf= v₀+at   v₀ = 0, vf = 25 m/s ,  t= 0.05 s

vf-v₀ = at

a =  (vf -v₀ ) /(t)

a = (25 -0)/(0.05)

a = 500 m/s²

We calculater the force exerted by the boxer on the sheet of paper (F) using formula (2):

F= m*a

F= 0.005 kg*500 m/s²

F = 2.5 N

get the total number of pieces of fruit and divide it by 4 to get the average. the result is the answer to the problem. 14 total number of fruits and there are 4 members. 14 divided by 4 is equals to 3.5.